X^2 - 4x - 4y = 0 in standard form

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

I took a picture of the question

cluponka [151]

Answer:

27.5 inches^2

Step-by-step explanation:

Area of a triangle =

$\frac{bh}{2}$

Plugging in the equation,

$\frac{11 \times 5}{2} = \frac{55}{2} \\ \frac{55}{2} = 27.5$

7 0

2 years ago

Which lines are parellele

natali 33 [55]

Line 1 and Line 4 are parallel lines

Solution:

General equation of a line:

y = mx + c

where m is the slope and c is the y-intercept of the line.

<u>To find the slope of each line:</u>

Line 1: $y=-\frac{3}{4} x-6$

Slope $(m_1)=-\frac{3}{4}$

Line 2: $y-7=-4(x+9)$

$y-7=-4x-36$

Add 7 on both sides, we get

$y=-4x-29$

Slope $(m_2)=-4$

Line 3: $y=-2 x+6$

Slope $(m_3)=-2$

Line 4: $x+\frac{4}{3} y=-5$

Subtract x from both sides, we get

$\frac{4}{3} y=-x-5$

Multiply by $\frac{3}{4}$ on both sides, we get

$y=-\frac{3}{4}x-\frac{15}{4}$

Slope $(m_4)=-\frac{3}{4}$

<em>Two lines are parallel, if their slopes are equal.</em>

From the above slopes,

$m_1=m_4$

Therefore Line 1 and Line 4 are parallel lines.

3 0

3 years ago

Can somebody help me with this math problem?

skad [1K]

Answer:

B) Rational, not an integer.

Step-by-step explanation:

The number is a decimal continuing forever; it is predictable. This means that it is Rational. If the number was not repeating then it would be irrational.

The number is not an integer because it is not a whole number; it is a decimal.

7 0

3 years ago

How do you solve this problem? (x+3) cubed

Lubov Fominskaja [6]

You would expand becausee it isn't equal to anything

(x+3)^3=(x+3)(x+3)(x+3)
we can use binomial theorem or pascal's triangle
anyway, pascal's triangle is more practicl for this one

so
for the 3rd power
$(a+b)^3=1a^3+3a^2b^1+3a^1b^2+1b^3$

a=x
b=3

$(x+3)^3=1x^3+3x^2(3^1)+3x^1(3)^2+1(3)^3$ =
$x^3+9x^2+27x+27$

7 0

3 years ago

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